Linear Algebra, subset of R2 not closed under scalar multipl

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Terrell
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Homework Statement


Construct a subset of the x-y plane R2 that is
(a) closed under vector addition and subtraction, but not scalar multiplication.

Hint: Starting with u and v, add and subtract for (a). Try cu and cv

Homework Equations


vector addition, subtraction and multiplication

The Attempt at a Solution


i think it's impossible for a combination of two matrices of rank two to have a matrix beyond rank 2. i can get it down to rank 1 and rank 0 of course but that is still a subset of R2 and that still proves that it's closed under scalar multiplication
 
on Phys.org
Consider the set of all vectors S = [x, y] such that x, y are integers.

Does that work?
 
Exus said:
Consider the set of all vectors S = [x, y] such that x, y are integers.

Does that work?
you mean xA + yB where A and B are rank 2 matrices?
 
fresh_42 said:
How come matrices into play?
Do you know what the vector subspaces of ##\mathbb{R}^2## are?
well R2 is a subspace of itself right? am i wrong. can anyone clear up any of my misunderstandings? thanks
 
fresh_42 said:
Yes. And what are the one-dimensional subspaces?
it's a line. and it's still under R2
 
the way i understand the question is. to construct a subset of R2 that is not within R2. therefore something in R3 or above?
 
fresh_42 said:
Yes. But a special kind of line.
but wouldn't that still be in R2?
 
fresh_42 said:
Your first sentence says "construct a subset of ##\mathbb{R}^2## ..."
a subset in R2 not closed under scalar multiplication... would that mean an R1 or R0? because R1 and R0 is still in R2 therefore, if the resulting matrix is still within R2, then it's closed under scalar multiplication... right?
 
Terrell said:
a subset in R2 not closed under scalar multiplication... would that mean an R1 or R0? because R1 and R0 is still in R2 therefore, if the resulting matrix is still within R2, then it's closed under scalar multiplication... right?
What do you mean by a matrix in this context? A subset that is closed under addition but not under scalar multiplication cannot be a subspace.
Therefore I asked for the subspaces, because these cannot be a solution. Any set of points ##(x,y)## is a subset.

Edit: Let's take one single point and look where addition and subtraction gets us to.
 
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Closed under scalar multiplication means that any vector in the subset could be multiplied by a scalar and still be within the subset.
 
fresh_42 said:
What do you mean by a matrix in this context? A subset that is closed under addition but not under scalar multiplication cannot be a subspace.
Therefore I asked for the subspaces, because these cannot be a solution. Any set of points ##(x,y)## is a subset.
Exus said:
Closed under scalar multiplication means that any vector in the subset could be multiplied by a scalar and still be within the subset.
i understood "closed under multiplication" correctly then. but i think what the question is asking is a subset not closed under multiplication that's why it's not making any sense to me because as fresh_42 said that in order to be a subspace, it must be closed under multiplication
 
There is a difference between subspace and subset. The first is a vector space itself (closed under addition and scalar multiplication) whereas a subset is just a set. But here we have to include all sums and differences, so it has to be an infinite set.

What happens if you take a certain point into your subset? You will have to include all sums and differences as well.
That's the smallest set you can get.
 
fresh_42 said:
There is a difference between subspace and subset. The first is a vector space itself (closed under addition and scalar multiplication) whereas a subset is just a set. But here we have to include all sums and differences, so it has to be an infinite set.

What happens if you take a certain point into your subset? You will have to include all sums and differences as well.
That's the smallest set you can get.
thank you for the clarification. i thought subspace = subsets, forgot that subsets doesn't necessarily have the zero vector
 
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Terrell said:
the way i understand the question is. to construct a subset of R2 that is not within R2. therefore something in R3 or above?
No, this makes no sense. Any subset of R2 is contained in R2. What you said here is probably due to your confusion about the difference between a subset of some space and a subspace of that space.
 
Mark44 said:
No, this makes no sense. Any subset of R2 is contained in R2. What you said here is probably due to your confusion about the difference between a subset of some space and a subspace of that space.
most definitely sir!